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1 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun

TWO PORT NETWORK

Port: It is a pair of terminals through which a current may enter or leave a network

Examples

Note: Reference direction and notations for port variable is shown in the figure (h).

Two-port Network Parameters:

S.No Name of Parameter Function Matrix Equation

1. Open Circuit Impedance Parameters or

Z-Parameters

( ) ( )2121 ,, IIfVV =

=

2

1

2221

1211

2

1

I

I

ZZ

ZZ

V

V

2. Short Circuit Admittance Parameters or

Y-Parameters

( ) ( )2121 ,, VVfII =

=

2

1

2221

1211

2

1

V

V

YY

YY

I

I

3. Transmission Parameters or

T-Parameters or A,B,C,D Parameters

( ) ( )2211 ,, IVfIV −=

−

=

2

2

1

1

I

V

DC

BA

I

V

4. Inverse Transmission Parameters or

T’-Parameters or A

’,B

’,C

’,D

’ Parameters

( ) ( )1122 ,, IVfIV −=

−

=

1

1

2

2

I

V

DC

BA

I

V

Fig .1

Network

(a) Two Port Network

Network

(c) One Port Network

Network

(b) Two Port Network

Network

(e) Three Port Network

Network

(d) Four Port Network

(g) T-Network(h) Π Network (f) T-Network

Network +

-

+

-

V1 V2

I1 I2

I1 I2

I/P Port O/P Port


5. Hybrid Parameters or

h parameters

( ) ( )2121 ,, VIfIV =

=

2

1

2221

1211

2

1

V

I

hh

hh

I

V

6. Inverse Hybrid Parameters or

g parameters

( ) ( )2121 ,, IVfVI =

=

2

1

2221

1211

2

1

I

V

gg

gg

V

I

Method for calculation of different parameters

1. Open Circuit Impedance Parameters or Z-Parameters:

We know for z-parameters

( ) ( )2121 ,, IIfVV =

=

2

1

2221

1211

2

1

I

I

ZZ

ZZ

V

V

)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

From above the equations the equivalent circuit will be

Case I: Output port is open circuited, i.e. I2 = 0

Put I2 = 0 in equation (1) & (2)

impedancepointdrivinginputcircuitopen

01

111

2

===I

I

VZ

impedancetransferforwardcircuitopen

01

221

2

===I

I

VZ

(b) Case I: I2 = 0

Network V1

I1 I2=0

V2

V1

I1

Z11

Z12I2 V2

I2

Z22

Z21I1

(a) Equivalent Circuit for Z-Parameters


Case II: Input port is open circuited, i.e. I1 = 0


mpedancetransferreversecircuitopen

02

112

1

iI

VZ

I

===

impedancepointdrivingoutputcircuitopen

02

222

1

===I

I

VZ

Alternate method:

Step 1: Apply KVL/KCL in the given circuit.

Step 2: Rearrange the above equations in the following forms

( )211 , IIfV =

( )212 , IIfV =

Step 3: Compare the above two equations with the standard equations of z-parameters.

Example 1: Determine Z-parameters for the network shown in figure 3a.

Solution: We know for z-parameters

)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

Case I: When I2 = 0, i.e. output port is open circuited, fig.3b

Ω==⇒=+= 1010551

1111111

I

VZIIIV

Ω==⇒= 551

22112

I

VZIV

Case II: When I1 = 0, i.e. input port is open circuited, fig.3c

(c) Case II: I1 = 0

Network V1

I1=0 I2

V2

Fig. 2

5Ω 5Ω V1

I1

V2

I2

(a)

5Ω5Ω V1

I1

V2

I2=0

(b)

5Ω 5Ω V1

I1=0

V2

I2

(c)

Fig.3


Ω==⇒= 552

11221

I

VZIV

Ω==⇒= 552

22222

I

VZIV

Hence

=

55

510

2221

1211

ZZ

ZZ Answer

Alternate Method:

Apply KVL in LHS loop of fig.3a

)3(510)(55 2112111 −−−−−+=⇒++= IIVIIIV

Apply KVL in RHS loop of fig.3a

)4(55)(5 212212 −−−−−+=⇒+= IIVIIV

Comparing equation (3) with equation (1) and equation (4) with equation (2)

=

55

510

2221

1211

ZZ

ZZ Answer

Note: Alternate method is easer, so we will apply only this method for rest of the examples in

these notes.

Example 2: Determine Z-parameters for the network shown in figure 4.


)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

Apply KVL in LHS loop

)3(3

2

3

42)(22 21112111 −−−−−+=⇒−++= IIVVIIIV

2Ω 2Ω V1

I1

V2

I2

Fig.4

2V1


Apply KVL in RHS loop

+−+=−+= 212112123

2

3

42)(22)(2 IIIIVIIV

)4(3

2

3

2212 −−−−−+−=⇒ IIV


−=

3

2

3

2

3

2

3

4

2221

1211

ZZ

ZZ Answer

2. Short Circuit Admittance Parameters or Y-Parameters:

We know for Y-parameters

( ) ( )2121 ,, VVfII =

=

2

1

2221

1211

2

1

V

V

YY

YY

I

I

)1(2121111 −−−−−+= VYVYI

)2(2221212 −−−−−+= VYVYI


Case I: Output port is short circuited, i.e. V2 = 0

Put V2 = 0 in equation (1) & (2)

admittancepointdrivinginputcircuitshort

01

111

2

===V

V

IY

V1

I1

Y11 Y12I2 V2

I2

Y22 Y21I1

(a) Equivalent Circuit for Y-Parameters

(b) Case I: V2 = 0

Network V1

I1 I2

V2=0


admittancetransferforwardcircuitshort

01

221

2

===V

V

IY

Case II: Input port is short circuited, i.e. V1 = 0


admittancetransferreversecircuitshort

02

112

1

===V

V

IY

admittancepointdrivingoutputcircuitshort

02

222

1

===V

V

IY

Alternate method:



( )211 ,VVfI =

( )212 ,VVfI =

Step 3: Compare the above two equations with the standard equations of y-parameters.

Example 3: Determine Z & Y-parameters for the network shown in figure 6a.


)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

And for Y-parameters

)3(2121111 −−−−−+= VYVYI

)4(2221212 −−−−−+= VYVYI

Fig. 5

(c) Case II: V1 = 0

Network V1=0

I1 I2

V2

1Ω 2Ω V1

I1

V2

I2

(a)

3I2

1

2 V1

I1

V2

I2

(b)

3I2

I1-3I2

I1-2I2

Fig.6


Apply KVL LHS

)2(2)3(1 21211 IIIIV −+−= )5(73 211 −−−−−−=⇒ IIV

Apply KVL RHS loop

)2(2 212 IIV −= )6(42 212 −−−−−−=⇒ IIV


−

−=

42

73

2221

1211

ZZ

ZZ Answer

Now rearrange equation (5) and (6) according to equation (3) & (4) respectively.

( ) ( ) 7645 ×−× )7(2

72 211 −−−−−+−=⇒ VVI

( ) ( ) 3625 ×−× )8(2

31 212 −−−−−+−=⇒ VVI


−

−=

2

31

2

72

2221

1211

YY

YY Answer

3. Transmission Parameters or T-Parameters or ABCD Parameters:

We know for T-parameters

( ) ( )2211 ,, IVfIV −=

−

=

2

2

1

1

I

V

DC

BA

I

V

)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI

Note: Equivalent circuit is not possible.


Put I2 = 0 in equation (1) & (2) (a) Case I: I2 = 0

Network V1

I1 I2=0

V2


ratiotransfervoltagereversecircuitopen

02

1

2

===I

V

VA

admittancetransferreversecircuitopen

02

1

2

===I

V

IC



impedancetransferreversecircuitshort

02

1

1

=−

==V

I

VB

ratiotransfercurrentreversecircuitshort

02

1

1

=−

==V

I

ID

Alternate method:



( )221 , IVfV =

( )221 , IVfI =

Step 3: Compare the above two equations with the standard equations of T-parameters.

4. InverseTransmission Parameters or T’-Parameters or A

’ B’ C’ D’ Parameters:

We know for T’-parameters

( ) ( )1122 ,, IVfIV −=

−

=

1

1

''

''

2

2

I

V

DC

BA

I

V

)1(1'

1

'

2 −−−−−−= IBVAV

)2(1'

1

'

2 −−−−−−= IDVCI

Note: Equivalent circuit is not possible.

Fig. 7

(b) Case II: V1 = 0

Network V1=0

I1 I2

V2


Case I: Input port is open circuited, i.e. I1 = 0


ratiotransfervoltagereversecircuitopen'

02

1

1

===I

V

VA

admittancetransferforwardcircuitopen

01

2'

1

===I

V

IC



impedancetransferforwardcircuitshort

01

2'

1

=−

==V

I

VB

ratiotransfercurrentforwardcircuitshort

01

2'

1

=−

==V

I

ID

Alternate method:



( )112 , IVfV =

( )112 , IVfI =

Step 3: Compare the above two equations with the standard equations of T’-parameters.

Example 4: Determine Z & T & T’-parameters for the network shown in figure 9a.

(a) Case I: I1 = 0

Network V1

I1=0 I2

V2

Fig. 8

(b) Case II: V1 = 0

Network V1=0

I1 I2

V2

1Ω 1Ω V1

I1

V2

I2

(a)

1Ω

2I12I2

V1

I1

V2

I2

(b)

2I12I2

I1+2I2 I2+2I1

2I1+2I2

I1+I2

1 Ω

1 Ω 1 Ω

Fig.9



)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV


)3(221 −−−−−−= BIAVV

)4(221 −−−−−−= DICVI


)5(1'

1

'

2 −−−−−−= IBVAV

)6(1'

1

'

2 −−−−−−= IDVCI

Two loop equations of fig.9b are

)7(32)(1)2(1 21121211 −−−−−+=⇒+++= IIVIIIIV

)8(23)(1)2(1 21221212 −−−−−+=⇒+++= IIVIIIIV

Comparing equation (1) & (2) with equation (7) & (8)

=

23

32

2221

1211

ZZ

ZZ Answer


( ) ( ) 2837 ×−× )9(3

5

3

2221 −−−−−+=⇒ IVV

( ) ( ) 3827 ×−× )10(3

2

3

1221 −−−−−−=⇒ IVI

Comparing equation (9) &(10) with equation (3) & (4) respectively

−=

3

2

3

1

3

5

3

2

DC

BA Answer



( ) ( ) 3827 ×−× )11(3

5

3

2112 −−−−−−=⇒ IVV

From equation (8) )12(3

2

3

1112 −−−−−−=⇒ IVI

Comparing equation (11) & (12) with equation (5) & (6) respectively

−=

3

2

3

1

3

5

3

2

''

''

DC

BA Answer

5. Hybrid Parameters or h-Parameters:

We know for h-parameters

( ) ( )2121 ,, VIfIV =

=

2

1

2221

1211

2

1

V

I

hh

hh

I

V

)1(2121111 −−−−−+= VhIhV

)2(2221212 −−−−−+= VhIhI


Case I: Output port is short circuited, i.e. V2 = 0


impedancepointdrivinginputcircuitshort

01

111

2

===V

I

Vh

admittancetransferforwardcircuitshort

01

221

2

===V

V

Ih

V1

I1

h11

h12V2

(a) Equivalent Circuit for h-Parameters

V2

I2

h22 h21V1

(b) Case I: V2 = 0

Network V1

I1 I2

V2=0


Case II: Input port is open circuited, i.e. I1 = 0


ratiotransfervoltage reversecircuit open

02

112

1

===I

V

Vh

admittancetransferforwardcircuitopen

02

222

1

===I

V

Ih

Alternate method:



( )121 ,VIfV =

( )122 ,VIfI =

Step 3: Compare the above two equations with the standard equations of h-parameters.

6. Inverse Hybrid Parameters or g-Parameters:


( ) ( )2121 ,, IVfVI =

=

2

1

2221

1211

2

1

I

V

gg

gg

V

I

)1(2121111 −−−−−+= IgVgI

)2(2221212 −−−−−+= IgVgV


(c) Case I: I1 = 0

Network V1

I1=0 I2

V2

Fig. 10

V1

I1

g11 g12I2

(b) Equivalent Circuit for g-Parameters

V2

I2

g22

g21V1




admittanceinputcircuitopen

01

111

2

===I

V

Ig

gainvoltageforwardcircuitopen

01

221

2

===I

V

Vg



gaincurrentreversecircuitshort

02

112

1

===V

I

Ig

impedanceoutputcircuitshort

02

222

1

===V

I

Vg

Alternate method:



( )211 , IVfI = & ( )212 , IVfV =

Step 3: Compare the above two equations with the standard equations of g-parameters.

Example 5: Determine Y & h & g-parameters for the network shown in figure 12a.

Solution: We know for Y-parameters

)1(2121111 −−−−−+= VYVYI

)2(2221212 −−−−−+= VYVYI

(b) Case I: I2 = 0

Network V1

I1 I2=0

V2

Fig. 11

(b) Case II: V1 = 0

Network V1=0

I1 I2

V2

Fig.12

1Ω 1Ω V1

I1

V2

I2

(a)

1Ω

1Ω 1Ω V1

I1

V2

I2

(b)

1Ω V1 V2



)3(2121111 −−−−−+= VhIhV

)4(2221212 −−−−−+= VhIhI

We know for g-parameters

)5(2121111 −−−−−+= IgVgI

)6(2221212 −−−−−+= IgVgV

Apply KCL at node V1 of fig.12b

)7(211

211211

1 −−−−−−=⇒−

+= VVIVVV

I

Apply KCL at node V2 of fig.12b

)8(2111

212221

2 −−−−−+−=⇒=−

+ VVIVVV

I

Comparing equation (1) & (2) with equation (7) & (8)

−

−=

21

12

2221

1211

YY

YY Answer



1

2

1211 −−−−−+= VIV

2)8()7( ×+ )10(2

3

2

1212 −−−−−+−= VII

Comparing equation (3) & 4) with equation (9) & (10)

−=

2

3

2

1

2

1

2

1

2221

1211

hh

hh Answer


)8(2)7( +× )11(2

1

2

3211 −−−−−−= IVI



1

2

1212 −−−−−+= IVV

Comparing equation (5) & 6) with equation (11) & (12)

−=

2

1

2

1

2

1

2

3

2221

1211

gg

gg Answer

Inter-relationship of parameters:

Since all the parameters describe the same two port network, they are inter related. One set of

parameters may be expressed in terms of the other set.

1. Z-Parameters:

We know for z-parameters

)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

(a) Z-Parameters in terms of Y-Parameters:


)3(2121111 −−−−−+= VYVYI

)4(2221212 −−−−−+= VYVYI

Now rearrange equation (3) & (4) according to equation (1) & (2) respectively

1222 )4()3( YY ×−× )5(212

122

1 −−−−−−=⇒ IY

YI

Y

YV

Where 211222112221

1211YYYY

YY

YYY −==

1121 )4()3( YY ×−× )6(211

121

1 −−−−−+−=⇒ IY

YI

Y

YV

Compering equation (1) & (2) with equation (5) & (6) respectively


−

−

=

y

Y

y

Y

y

Y

y

Y

ZZ

ZZ

1121

1222

2221

1211

Note: [ ] 121221

1222

1121

1222

2221

1211 1 −=

−

−=

−

−

=

Y

YY

YY

Y

y

Y

y

Y

y

Y

y

Y

ZZ

ZZ

So [ ] [ ] 1−= YZ

(b) Z-Parameters in terms of T-Parameters:


)3(221 −−−−−−= BIAVV

)4(221 −−−−−−= DICVI


BD ×−× )4()3( )5(211 −−−−−+=⇒ IC

TI

C

AV

Where CBADDC

BAT −==

equ(3)From )6(1

212 −−−−−+=⇒ IC

DI

CV


=

C

D

C

C

T

C

A

ZZ

ZZ

12221

1211


(c) Z-Parameters in terms of T’-Parameters:


)3(1'

1

'

2 −−−−−−= IBVAV

)4(1'

1

'

2 −−−−−−= IDVCI


equ(4)From )6(1

2'1'

'

1 −−−−−+=⇒ IC

IC

DV

'' )4()3( AC ×−× )5(2'

'

1'

'

2 −−−−−+=⇒ IC

AI

C

ATV

Where ''''''

''

' BCDADC

BAT −==


=

'

'

'

'

''

'

2221

1211

1

C

A

C

T

CC

D

ZZ

ZZ

(d) Z-Parameters in terms of h-Parameters:


)3(2121111 −−−−−+= VhIhV

)4(2221212 −−−−−+= VhIhI


1222 )4()3( hh ×−× )5(222

121

22

1 −−−−−+=⇒ Ih

hI

h

hV


2

22

1

22

212 −−−−−+−=⇒ I

hI

h

hV



−

=

2222

21

22

12

22

2221

1211

1

hh

h

h

h

h

h

ZZ

ZZ

(e) Z-Parameters in terms of g-Parameters:

We know for g-parameters

)3(2121111 −−−−−+= IgVgI

)4(2221212 −−−−−+= IgVgV



2

11

121

11

1 −−−−−−−=⇒ Ig

gI

gV

1121 )4()3( gg ×−× )6(211

1

11

212 −−−−−−+=⇒ I

g

gI

g

gV


−

=

1111

21

11

12

11

2221

1211

1

g

g

g

g

g

g

g

ZZ

ZZ

Example 6: Calculate values of Z-parameters, if the values of other parameters are given bellow:

(a) A = 2, B = -1, C = 3, D = -2

(b) h11 =1, h12 = -2, h21 = -3 h22 = 2

(c) Y11 = 1/3, Y12 = 2/3, Y21 = -1/3, Y22 =1/6

Solution: 1−=−= BCADT

Answer

−

−

=

=

3

2

3

1

3

1

3

2

12221

1211

C

D

C

C

T

C

A

ZZ

ZZ


421122211 −=−= hhhhh

−−=

−−

−−

=

−

=

2/12/3

12

2

1

2

3/1

2

2

2

4

1

2222

21

22

12

22

2221

1211

hh

h

h

h

h

h

ZZ

ZZ Answer

18/521122211 =−= YYYYY

−=

−

−

=

5

6

5

6

5

12

5

3

1121

1222

2221

1211

y

Y

y

Y

y

Y

y

Y

ZZ

ZZ Answer

2. Y-Parameters: Similar to above

(a) In terms of Z-Parameters:

(b) In terms of T-Parameters:

(c) In terms of T’-Parameters:

(d) In terms of h-Parameters:

(e) In terms of g-Parameters:

3. T-Parameters: Similar to above


(b) In terms of Y-Parameters:

(c) In terms of T’-Parameters:




4. T’-Parameters: Similar to above



(c) In terms of T-Parameters:



5. h-Parameters: Similar to above




(d) In terms of T’-Parameters:


6. g-Parameters: Similar to above




(d) In terms of T’-Parameters:

(e) In terms of h-Parameters:


Relationships between parameters:

Note:-

1. [ ] [ ] 1−= YZ & [ ] [ ] 1−= ZY

2. [ ] [ ] 1' −≠ TT & [ ] [ ] 1' −≠ TT

3. [ ] [ ] 1−= gh & [ ] [ ] 1−= hg


T & Π -Network:

Z-parameters of a T-network Y-parameters of a network−Π

ca ZZZ +=11 ca YYY +=11

cb ZZZ +=22 cb YYY +=22

cZZZ == 2112 cYYY −== 2112

T to Π& Π to T Transformation:

T to Π tionstransformadeltatostar⇒ and Π to T tionstransformastartodelta⇒

T to Π (Star to Delta) Π to T (Delta to Star)

b

accbba

a

Y

YYYYYYZ

1

111111

+

+

= cba

ca

a ZZZ

ZZ

Y ++=

1

a

accbba

b

Y

YYYYYYZ

1

111111

+

+

= cba

cb

b ZZZ

ZZ

Y ++=

1

c

accbba

c

Y

YYYYYYZ

1

111111

+

+

= cba

ab

c ZZZ

ZZ

Y ++=

1

Za

V1

I1

V2

I2

(a)

Zb

ZcYa V1

I1

V2

I2

(b)

Yb

Yc

Fig.13


Condition for reciprocity & reciprocal network:

A two port network is said to be reciprocal, if the ration of the excitation to response is invarient

to an interchage of the exciation and response in the network. Mathematicall, we can say from

following figure

01

2

02

1

12

if

==

=VV

I

V

I

V

0

'

10

'

212 ==

−=

−⇒

V

s

V

s

I

V

I

V

NetworkRecoprocal'2'

1

0

'

10

'

212

⇒==⇒==

IIORI

V

I

V

V

s

V

s

1. Reciprocity in terms of Z-Parameters:

We know for Z-parameters

)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

From fig.14a V1=Vs, I1=I1, I2=-I2’, V2=0 Put these conditions in equation (1), (2)

)3(0 21

21122211

'

2'

222121

'

221111 −−−−−−

=⇒

−=

−=

Z

ZZZZ

I

V

IZIZ

IZIZVss

From fig.14b V1=0, I1=-I1’, I2= I2, V2=Vs Put these conditions in equation (1), (2)

)4(0

12

21122211

'

1222

'

121

221

'

111 −−−−−−

=⇒

+−=

+−=

Z

ZZZZ

I

V

IZIZV

IZIZs

s

Fig. 14(V1=Vs, I1=I1, I2=-I2’, V2=0)

(a)

Network Vs

I1

'

2I V2=0

I2

Network V1=0

'

1I

I2

Vs

(V1=0, I1=-I1’, I2= I2, V2=Vs)

(b)

I1


But for reciprocal network

21120

'

10

'

212

ZZI

V

I

V

V

s

V

s =⇒===

2. Reciprocity in terms of Y-Parameters:


)1(2121111 −−−−−+= VYVYI

)2(2221212 −−−−−+= VYVYI


} )3(21'2

21

'

2

111

−−−−−−=⇒=−

=

YI

VVYI

VYI

ss

s


}

s

ss

VYI

YI

VVYI

222

12'

1

12

'

1 )4(

=

−−−−−−=⇒=−


21120

'

10

'

212

YYI

V

I

V

V

s

V

s =⇒===

3. Reciprocity in terms of T-Parameters:


)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI

2112Hence YY =

2112Hence ZZ =



}'

21

'

2

'

2 )3(

DII

BI

VBIV ss

=

−−−−−=⇒=


)4(0

'

12

'

1

2 −−−−−−

=⇒

−=−

−=

BCAD

B

I

V

DICVI

BIAVs

s

s


11

0

'

10

'

212

==−⇒===

DC

BAorBCAD

I

V

I

V

V

s

V

s

4. Reciprocity in terms of T’-Parameters: Similar to above

5. Reciprocity in terms of h-Parameters:


)1(2121111 −−−−−+= VhIhV

)2(2221212 −−−−−+= VhIhI


)3(21

11

'

2121

'

2

111−−−−−−=⇒

=−

=

h

h

I

V

IhI

IhVss


1Hence''

''

=DC

BA

1Hence =DC

BA


)4(0

12

11

'

122

'

1212

12

'

111 −−−−−=⇒

+=

+=

h

h

I

V

VhIhI

VhIhs

s

s


2112

0

'

10

'

212

hhI

V

I

V

V

s

V

s −=⇒===

6. Reciprocity in terms of g-Parameters: Similar to above

Condition for symmetry & Symmetrical networks:

A two port network is said to be symmertical if the ports can be interchaged without changing

the port voltages and currents. Mathematicall, we can say from following figure

NetworklSymmetricaif

020102

2

01

1

1212

⇒=⇒===== I

s

I

s

III

V

I

V

I

V

I

V

1. Symmetry in terms of Z-Parameters:


)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

From fig.15a V1=Vs, I1=I1, I2=0, V2=V2 Put these conditions in equation (1)

111IZVs =

2112Hence gg −=

2112Hence hh −=

Fig. 15(V1=Vs, I1=I1, I2=0, V2=V2)

(a)

Network Vs

I1 I2=0

V2 Network

V1

I1=0 I2

Vs

(V1=V1, I1=0, I2= I2, V2=Vs)

(b)


11

0101

1

22

ZI

V

I

V

I

s

I

==⇒==

From fig.15b V1=V1, I1=0, I2=I2, V2=Vs Put these conditions in equation (2)

222IZVs =

22

0202

2

11

ZI

V

I

V

I

s

I

==⇒==

221102

2

01

1

12

if ZZI

V

I

V

II

=====

2. Symmetry in terms of Y-Parameters:


)1(2121111 −−−−−+= VYVYI

)2(2221212 −−−−−+= VYVYI

From fig.15a V1=Vs, I1=I1, I2=0, V2=V2 Put these conditions in equation (1) & (2)

212111 VYVYI s +=

222210 VYVY s +=

21122211

22

0101

1

22

YYYY

Y

I

V

I

V

I

s

I−

==⇒==

From fig.15b V1=V1, I1=0, I2=I2, V2=Vs Put these conditions in equation (1) & (2)

sVYVY 121110 +=

sVYVYI 221212 +=

21122211

11

0202

2

11

YYYY

Y

I

V

I

V

I

s

I−

==⇒==

2211Hence ZZ =


2211

02

2

01

1

12

if YYI

V

I

V

II

=====

3. Symmetry in terms of T-Parameters:


)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI


2AVVs =

21 CVI =

C

A

I

V

I

V

I

s

I

==⇒== 0101

1

22


21 BIAVV s −=

20 DICVs −=

C

D

I

V

I

V

I

s

I

==⇒== 0202

2

11

DAI

V

I

V

II

===== 02

2

01

1

12

if

4. Symmetry in terms of T’-Parameters: Similar to above

''Hence DA =

DA =Hence

2211Hence YY =


5. Symmetry in terms of h-Parameters:


)1(2121111 −−−−−+= VhIhV

)2(2221212 −−−−−+= VhIhI


212111 VhIhVs +=

2221210 VhIh +=

22

21122211

0101

1

22

h

hhhh

I

V

I

V

I

s

I

−==⇒

==


sVhV 121 =

sVhI 222 =

220202

2 1

11

hI

V

I

V

I

s

I

==⇒==

11if2221

1211

21122211

02

2

01

1

12

==−⇒===

hh

hhorhhhh

I

V

I

V

II

6. Symmetry in terms of g-Parameters: Similar to above

1Hence2221

1211 =gg

gg

1Hence2221

1211 =hh

hh


Summary table for reciprocal and symmetrical two port network:-

S.No Parameters Condition of Reciprocity Condition of Symmetry

1. [ ]Z 2112 ZZ = 2211 ZZ = 2. [ ]Y

2112 YY = 2211 YY =

3. [ ]T 1=

DC

BA

DA =

4. [ ]'T 1

''

''

=DC

BA

'' DA =

5. [ ]h 2112 hh −= 1

2221

1211 =hh

hh

6. [ ]g 2112 gg −= 1

2221

1211 =gg

gg

Example 7: Calculate values of Z-parameters, and check whether given network is reciprocal

and symmetrical or not.

Solution: Take the Laplace Transformation (s-domain) of above circuit and convert it into

equivalent T-circuit. See fig.16b

Where

Ω= 2aZ

Ω+

=

=ss

Zc81

2

4

12

( ) Ω+= sZb 24

2 Ω

4F

(a)

2 Ω

4 Ω 2H ZaV1

I1

V2

I2

(b)

Zb

Zc

Fig.16


We know Z-parameters of a T-network

s

sZZZ ca

81

16411 +

+=+=

s

ssZZZ cb

81

16346 2

22 +++

=+=

s

ZZZ c81

22112 +

===

Hence [ ]

+++

++

++

++

=

s

ss

s

s

s

s

s

s

Z

81

16346

81

164

81

164

81

164

2 Answer

Clearly lSymmetricaNot2211 ⇒≠ ZZ

Reciprocal2112 ⇒= ZZ


Interconnections of two port networks:

1. Series connection or series-series connection:

Z-parameters for network A Z-parameters for network B

=

a

a

aa

aa

a

a

I

I

ZZ

ZZ

V

V

2

1

2221

1211

2

1

=

b

b

bb

bb

b

b

I

I

ZZ

ZZ

V

V

2

1

2221

1211

2

1

From above fig.17

ba III 111 == & ba III 222 ==

ba VVV 111 += & ba VVV 222 +=

So

+

=

+

+=

b

b

a

a

ba

ba

V

V

V

V

VV

VV

V

V

2

1

2

1

22

11

2

1

+

=

b

b

bb

bb

a

a

aa

aa

I

I

ZZ

ZZ

I

I

ZZ

ZZ

2

1

2221

1211

2

1

2221

1211

+

=

2

1

2221

1211

2

1

2221

1211

I

I

ZZ

ZZ

I

I

ZZ

ZZ

bb

bb

aa

aa

(Using ba III 111 == & ba III 222 == )

++

++=

2

1

22222121

12121111

2

1

I

I

ZZZZ

ZZZZ

V

V

aaaa

baba

Fig.17


So

++

++=

aaaa

baba

ZZZZ

ZZZZ

ZZ

ZZ

22222121

12121111

2221

1211

Or [ ] [ ] [ ]ba ZZZ +=

2. Parallel connection or Parallel-Parallel connection:

Y-parameters for network A Y-parameters for network B

=

a

a

aa

aa

a

a

V

V

YY

YY

I

I

2

1

2221

1211

2

1

=

b

b

bb

bb

b

b

V

V

YY

YY

I

I

2

1

2221

1211

2

1

From above fig.18

ba III 111 += & ba III 222 +=

ba VVV 111 == & ba VVV 222 ==

So

+

=

+

+=

b

b

a

a

ba

ba

I

I

I

I

II

II

I

I

2

1

2

1

22

11

2

1

+

=

b

b

bb

bb

a

a

aa

aa

V

V

YY

YY

V

V

YY

YY

2

1

2221

1211

2

1

2221

1211

+

=

2

1

2221

1211

2

1

2221

1211

V

V

YY

YY

V

V

YY

YY

bb

bb

aa

aa

Fig.18


(Using ba VVV 111 == & ba VVV 222 == )

++

++=

2

1

22222121

12121111

2

1

V

V

YYYY

YYYY

I

I

aaaa

baba

So

++

++=

aaaa

baba

YYYY

YYYY

YY

YY

22222121

12121111

2221

1211

Or [ ] [ ] [ ]ba YYY +=

3. Cascade Connection:

T-parameters for network A T-parameters for network B

−

=

a

a

aa

aa

a

a

I

V

DC

BA

I

V

2

2

1

1

−

=

b

b

bb

bb

b

b

I

V

DC

BA

I

V

2

2

1

1

From above fig.19

aII 11 = & ba II 12 =− & 22 II b =

aVV 11 = & ba VV 12 = & 22 VV b =

Fig.19


So

=

−

=

=

b

b

aa

aa

a

a

aa

aa

a

a

I

V

DC

BA

I

V

DC

BA

I

V

I

V

1

1

2

2

1

1

1

1

−

=

b

b

bb

bb

aa

aa

I

V

DC

BA

DC

BA

2

2

−

=

2

2

1

1

I

V

DC

BA

DC

BA

I

V

bb

bb

aa

aa

So

=

bb

bb

aa

aa

DC

BA

DC

BA

DC

BA

Or [ ] [ ][ ]ba TTT =

Similarly we can also drive a relation for T’-parameters

[ ] [ ][ ]''' ba TTT =

4. Series-parallel connection:

Similar to above

[ ] [ ] [ ]ba hhh +=

Fig.20


5. Parallel-series connection:

Similar to above

[ ] [ ] [ ]ba ggg +=

Overall parameters for different connections:

S.No Interconnection Overall Parameter

1. Series or Series-series [ ] [ ] [ ]ba ZZZ +=

2. Parallel or Parallel-parallel [ ] [ ] [ ]ba YYY +=

3. Cascade or Tandem [ ] [ ][ ]ba TTT =

4. Series-Parallel [ ] [ ] [ ]ba hhh +=

5. Parallel-Series [ ] [ ] [ ]ba ggg +=

Fig.21


Example 8: Determine the Y-parameters for the twin-T network shown in following figure, if all

resistances are of 1 ohm. (Fig.22)

Solution: Clearly above circuit is the parallel connection of two T-networks

We know Z-parameters of a T-network

[ ]

+

+=

cbc

cca

ZZZ

ZZZZ

So [ ] [ ]

==

21

12ba ZZ [ ] [ ] [ ]

−

−=

===⇒

−−

3

2

3

1

3

1

3

2

21

121

1

aba ZYY

[ ] [ ] [ ]

−

−=+=

3

4

3

2

3

2

3

4

ba YYY Answer

Example 9: Determine the transmission parameters for the network shown in following figure

using the concept of interconnection of four two-port networks N1, N2, N3 and N4 in cascade.

Solution:

We can directly write T-parameters of different network

=

10

211T

=

12

012

sT

=

10

213

sT

+=

11

014

sT

Fig.22

Fig.23

2 Ω

2F

2H

1F 1 Ω

N1 N2 N3 N4


So required T-parameters of given network

[ ] [ ][ ][ ][ ]

+

==

11

01

10

21

12

01

10

214321

s

s

sTTTTT

[ ]

++++

+++++=

141344

22838108

223

223

ssss

sssssT Answer

Example 10: Find out Z-parameters of following network.

Solution: We know for Z-parameters

)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

Apply KVL in loop 1-2’-1’ of fig.24b

( ) ( ) )3(132311 −−−−−−+−−= IIIZIIZV dc

Apply KVL in loop 2-1’-2’ of fig.24b

( ) ( ) )4(132322 −−−−−−+++= IIIZIIZV db

Voltage at 1-1 terminals via 1-2-1’= Voltage at 1-1 terminals via 1-2’-1’

( ) ( ) ( ) )5(13231323 −−−−−−++−=++ IIIZIIZIIZIZ dcba

Eliminate I3 from equations (1), (2) & (3) and rearrange them according to equations (1)

& (2)

( )( )

)6(211 −−−−−

+++

−+

+++

++= I

ZZZZ

ZZZZI

ZZZZ

ZZZZV

dcba

dacb

dcba

dcba

( )( ))7(212 −−−−−

+++++

+

+++−

= IZZZZ

ZZZZI

ZZZZ

ZZZZV

dcba

dbca

dcba

dacb

Za

Zb

Zc

Zd

(a) Fig.24

1

1’

2

2’

Za

Zb

Zc

Zd

(b)

V1 V2

I1 I2 I3

I1-I3

I2+I3

I1 I2+I3-I1 I2


Comparing equations (6) & (7) with equations (1) & (2)

[ ]

( )( )

( )( )

+++

++

+++

−

+++

−

+++

++

=

dcba

dbca

dcba

dacb

dcba

dacb

dcba

dcba

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

Z

For symmetrical network, i.e. cbda ZZZZ == &

[ ]

( ) ( )

( ) ( )

+−

−+

=

22

22

baab

abba

ZZZZ

ZZZZ

Z

Example 11: Find out Z-parameters of following network.

Solution:

Apply KVL at V1

)1(211 −−−−−+= VZIV

Apply KVL at V2

)2(212 −−−−−+

=Y

IIV

Rearranging these two equations, we have

( ) )3(.1 221 −−−−−−+= IZVZYV

)4(.1 221 −−−−−−= IYVI

Comparing equations (3) & (4) with standard T-parameters equations, we have

[ ]

+=

1

1

Y

ZZYT

Z

Y

Fig.25

V1 V2

I1 I2


Lattice Network: A two-port lattice network is shown in fig.31a. It can be simplified into

equivalent bridge network as shown in fig.31b. This is very common network used in filter

sections and also used as attenuator.

Z-parameters can be find out of a lattice network (See Example10)

[ ]

( )( )

( )( )

+++

++

+++

−

+++

−

+++

++

=

dcba

dbca

dcba

dacb

dcba

dacb

dcba

dcba

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

ZZZZ

Z

For symmetrical lattice network, i.e. cbda ZZZZ == &

[ ]

( ) ( )

( ) ( )

−−−=+=

−−−=−=⇒

+−

−+

=)2(

)1(

22

22

1211

1211

cb

da

baab

abba

ZZZZ

ZZZZ

ZZZZ

ZZZZ

Z

Hence if Z-parameters of a two-port network are known, we can find out its equivalent

lattice network using above equations (1) & (2)

Ladder Network: A two-port ladder network is shown in fig.27. It is mode of cascaded infinite

L-network. It is used in network synthesis.

T-parameters can be find out of a L-network (See Example11)

Za

Zb

Zc

Zd

(b)Bridge Equivalent N/W of Lattice N/W

2 2’

V2V1

1

1’

2

2’

V2V1

1

1’

Za

Zb

Zc

Zd

(a)Lattice Network Fig.26

Z

YV1

I1

Z

Y

Fig.27

Z

Y

Z

Y V2

I2


[ ] [ ] [ ] [ ] [ ]

+=====

1

1......4321

Y

ZZYTTTTT n

So T-parameters of a ladder network

[ ] [ ][ ][ ][ ] [ ]n

nY

ZZYTTTTTT

+==

1

1.......4321

Example 12: Obtain the lattice equivalent of a symmetrical T- network as shown in Fig28a.

Solution: A symmetrical and reciprocal two port network can be realized as a symmetrical two

port lattice network.

Z-parameters of above T-network

[ ] Reciprocal&lSymmetirca32

23⇒

=Z

We know for lattice network

523

123

1211

1211

=+=+==

=−=−==

ZZZZ

ZZZZ

cb

da

Equivalent lattice network is shown in fig28b

Open circuit and short circuit Impedance:

Open circuit impedance

( ) ( )2202

2211

01

11

12

& ZI

VZZ

I

VZ

I

o

I

o ≡=≡===

Short circuit impedance

≡=

≡=

== 2202

22

1101

11

1&

1

12

YI

VZ

YI

VZ

V

s

V

s

(a)

1 Ω 1 Ω 2 Ω

1 Ω

(b) Equivalent Lattice Network

1 Ω

5 Ω

5 Ω

Fig.28


1. Open circuit and short circuit impedances in terms of T-parameters:

We know T-parameter:

)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI

So )4(&)3(

02

22

01

11

12

−−−==−−−====

C

D

I

VZ

C

A

I

VZ

I

o

I

o

)6(&)5(

02

22

01

11

12

−−−==−−−====

A

B

I

VZ

D

B

I

VZ

V

s

V

s

2. T-parameters in terms of Open circuit and short circuit impedances:

From above equations (3), (4), (5) & (6)

so

o

ZZ

ZA

22

1

−±=

so

os

ZZ

ZZB

22

12 −

±=

( )soo ZZZ

C221

1

−±=

( )soo ZZZ

ZD221

20

1

−±=

Z1o 2-2’ is open circuited

Impedance at 1-1’

Z1s 2-2’ is short circuited

Impedance at 1-1’

Note:


Input and output Impedances:

Input impedance: If a load impedance ZL is connected across the output port, the impedance at

input port is known as input impedance,

1

1

I

VZin =

(i) Input impedance in terms of Z-parameters:


)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

From fig.29 )3(22 −−−−−−= LZIV

Put the value of V2 from equation (2) to equation (3)

)4(122

2122222121 −−−−−+

−=⇒−=+ I

ZZ

ZIZIIZIZ

L

L

Put the value of I2 from equation (4) to equation (1)

+

−+= 1

22

21121111 I

ZZ

ZZIZV

L

So

(ii) Input impedance in terms of T-parameters:


)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI

From fig.29 )3(22 −−−−−−= LZIV

Fig.29

Network V1

I1 I2

V2 ZL

L

Lin

ZZ

ZZZZZZ

I

VZ

++−

==22

1121122211

1

1


Put the value of V2 from equation (3) to equation (1) & equation (2)

( ) 221 BIZIAV L −−=

( ) 221 DIZICI L −−=

From above two equations

(iii) Input impedance in terms of h-parameters:


)1(1 212111

−−−−−+= VhIhV

)2(2221212 −−−−−+= VhIhI

From fig.29 )3(22 −−−−−−= LZIV


( ) )4(1

1

22

2122221212 −−−−−+

−=⇒+−= I

Zh

ZhVZVhIhV

L

LL


+−

+= 122

21121111

1I

Zh

ZhhIhV

L

L

So

Output impedance: If a load impedance ZL is connected across the input port, the impedance at

output port is known as output impedance,

2

2

I

VZop =

Fig. 30

Network V1

I1 I2

V2 ZL

DCZ

BAZ

I

VZ

L

Lin +

+==

1

1

( )L

Lin

Zh

hZhhhh

I

VZ

22

1121122211

1

1

1++−

==


(i) Output impedance in terms of Z-parameters:


)1(2121111 −−−−−+= IZIZV

)2(2221212 −−−−−+= IZIZV

From fig.30 )3(11 −−−−−−= LZIV


)4(211

2112121111 −−−−−+

−=⇒+=− I

ZZ

ZIIZIZZI

L

L


+−

+= 111

21221212 I

ZZ

ZZIZV

L

So

(ii) Input impedance in terms of T-parameters:


)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI

From fig.30 )3(11 −−−−−−= LZIV

Put the value of V1 & I1 from equation (1) & (2) to equation (3)

( ) LZDICVBIAV 2222 −−=−

So

L

Lop

ZZ

ZZhZZZ

I

VZ

+

+−==

11

2221122211

2

2

ACZ

BDZ

I

VZ

L

Lop +

+==

2

2


(iii) Input impedance in terms of h-parameters:


)1(1 212111

−−−−−+= VhIhV

)2(2221212 −−−−−+= VhIhI

From fig.30 )3(11 −−−−−−= LZIV


)4(211

1212121111 −−−−−+

−=⇒+=− V

Zh

hIVhIhVI

L

L


VhVZh

hhI

L

222

11

12212 +

+

−=

So

Image Impedances:

In a two port network, if the impedance at input port with impedance Zi2 connected at output port be Zi1 and the impedance at output port with impedance Zi1 connected

across input port be Zi2 as shown in following figure. Then these impedances Z1i & Z2i are

known as image impedances.

From fig.31a

portinputatimpedancepointDriving1

11 ==

I

VZ i

L

Lop

Zhhhhh

Zh

I

VZ

2221122211

11

2

2

+−

+==

(b)

Network V1

I1 I2

V2 Zi1

(a)

Network V1

I1 I2

V2 Zi2

Fig.31

Zi1→ ←Zi2


From fig.31b

portoutputatimpedancepointDriving2

22 ==

I

VZ i

For symmetrical network

impedancesticCharacteri21 === Cii ZZZ

(i) Image Impedance in terms of Input and output Impedance:

ipi ZZ =1 and opi ZZ =2

(ii) Image Impedance in terms of T-Parameters:

)1(2

21 −−−−−+

+==

DCZ

BAZZZ

i

iipi

)2(1

12 −−−−−+

+==

ACZ

BDZZZ

i

iopi

Solving equations (1) & (2)

CD

ABZ i =1 and

CA

DBZ i =2

(iii)Image Impedance in terms of Open circuit and short circuit impedances:

We know

soi ZZD

B

C

A

CD

ABZ 111 =×==

And soi ZZA

B

C

D

CA

DBZ 222 =×==

Note: The image impedances don’t completely define a two port network. So we need another

parameter which we shall get from the voltage and current rations as follows:


)1(221 −−−−−−= BIAVV

)2(221 −−−−−−= DICVI


And reversedisIparametersTIn:Note)3( 2222 −−−−−−−= iZIV

From equation (3) 2

22

iZ

VI −= put in equation (1)

222

221 V

Z

BA

Z

VBAVV

ii

+=

+=

)4(22

1 −−−−−

+=⇒

iZ

BA

V

V

Again from equation (3) put the value of V2 to equation (2)

( ) ( ) 222221 IDCZDIZICI ii +−=−−=

( ) )5(22

1 −−−−−+−=⇒ DCZI

Ii

But AC

BDZ i =2 put in equation (4) and (5)

)6(2

1 −−−−−+=+=D

ABCDA

AC

BDBA

V

V

And )7(2

1 −−−−−−+=+=−A

ABCDDD

AC

BDC

I

I

From above equation (6) and (7)

+

+=×−

A

ABCDD

D

ABCDA

I

I

V

V

2

1

2

1

BCABCDAD ++= 2

( )2BCAD +=

So 122

11 −+=+=− ADADBCADIV

IV 1=− BCADQ

Let θcosh=AD


So θsinh1 =−AD

Therefore θθθ eIV

IV=+=− sinhcosh

22

11

Or constanttransferImageln2

1ln

22

11

22

11 =

−=−=

IV

IV

IV

IVθ

Again 222111 & IZVIZV ii −== , the image transfer constant may be written as

=

=

2

22

2

11

222

111 ln2

1ln

2

1

IZ

IZ

IIZ

IIZ

i

i

i

iθ

+

=

+

=

2

1

2

1

2

2

2

1

2

1 lnln2

1lnln

2

1

I

I

Z

Z

I

I

Z

Z

i

i

i

i

Other form of image transfer constant

AD1cosh−=θ

1sinhsinh 11 −== −− ADBCθ

o

s

o

s

Z

Z

Z

Z

AD

BC

2

21

1

111 tanhtanhtanh −−− ===θ

Let 1

1tanh

2

2

1

1

+−

=+−

=== −−

θ

θ

θθ

θθ

θe

e

ee

ee

Z

Zm

o

s

⇒ φθ jrem

me =

−+

=1

12 ( )1o1s Z&Zfromcalculatedbecanr,φ

Hence ( ) βαφπθ jnjr +=

++=2

ln2

1 n=1, 2, 3…….etc

( ) constantnattenuatioImageln2

1== rα

constantphase2

=

+=φ

πβ n

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